Prove that 1/(1-tanx) - 1/(1+tanx) = tan2x .
To add the ratios, we'll have to have the same denominator. We'll calculate the least common denominator for the given ratios.
LCD = (1-tanx)(1+tanx)
We notice that the LCD is a difference of square:
LCD = 1 - (tan x)^2
To obtain the denominator 1 - (tan x)^2, we'll multiply the first ratio by (1+tanx) and the second ratio, by (1-tanx).
We'll re-write the left side:
1/(1-tanx) - 1/(1+tanx) = (1 + tan x - 1 + tan x)/[1 - (tan x)^2]
We'll eliminate and combine the like terms from numerator:
1/(1-tanx) - 1/(1+tanx) = 2tan x/[1 - (tan x)^2] (1)
Now, we'll re-write the right side:
tan 2x = tan (x+x)
We'll apply the formula for the tangent of the sum of 2 angles:
tan (x+x) = (tan x + tan x)/(1-tan x*tan x)
tan 2x = 2tan x/[1 - (tan x)^2] (2)
We notice that we have obtained (1) = (2), so the identity is verified, fro any value of x:
1/(1-tanx) - 1/(1+tanx) = 2 tan x